Circuits Help

C = 7.20 micro-Farads, R1 = 10.5ohm, R2 = 22.5ohm, V0 = 16.0V. What is the current (to within 0.01A) through R1 immediately after the switch S is closed? I created two equations to represent the voltage drops in the system but Capacitors charge up over time which is confusing me:

16V - i1*R1 - i2*r2 = 0
16V - i1R1 - q/C = 0

Therefore i2*R2 = q/C subtracting the two. Unfortunately that's two unknowns. I also sort of (but not fully understand) that for a resitor and capacitor in series, q = C*Vapplied*(1-e^(-t*R*C)) so at t = 0, q = 0 and thus no voltage drop across the capacitor and therefore no current. This means that all of the current is flowing through R2 right? Therefore i2 = i1 so..

16V - I1(R1+R2) = 0
16V - I1(10.5+22.5) =0
I = 0.48A?

Am I right to apply this to this drawing? Edit 1.33 was wrong. So was 0.48. OK I'm really lost now.

New theory: Is the capacitor being initially uncharged short-circuiting the system? Then the voltage drop along R1 would be 16V meaning that 16=10.5*I, I = 1.523A. Am I any closer? Edit YESS I got it :) Maybe there's hope for me yet.